# Restricted Measure is Measure

## Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.

Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\left({X, \Sigma'}\right)$.

## Proof

Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$:

### Axiom $(1)$

The statement of axiom $(1)$ for $\mu \restriction_{\Sigma'}$ is:

$\forall E' \in \Sigma': \mu \restriction_{\Sigma'} \left({E'}\right) \ge 0$

Now, for every $E' \in \Sigma'$, compute:

 $\displaystyle \mu \restriction_{\Sigma'} \left({E'}\right)$ $=$ $\displaystyle \mu \left({E'}\right)$ Definition of $\mu \restriction_{\Sigma'}$ $\displaystyle$ $\ge$ $\displaystyle 0$ $\mu$ is a measure

$\Box$

### Axiom $(2)$

Let $\left({E'_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma'$.

Then the statement of axiom $(2)$ for $\mu \restriction_{\Sigma'}$ is:

$\displaystyle \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right) = \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)$

One can show this by means of the following computation:

 $\displaystyle \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right)$ $=$ $\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E'_n}\right)$ Definition of $\mu \restriction_{\Sigma'}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop \in \N} \mu \left({E'_n}\right)$ $\mu$ is a measure $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)$ Definition of $\mu \restriction_{\Sigma'}$

$\Box$

### Axiom $(3')$

The statement of axiom $(3')$ for $\mu \restriction_{\Sigma'}$ is:

$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = 0$

By Sigma-Algebra Contains Empty Set, $\varnothing \in \Sigma'$. Hence:

$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = \mu \left({\varnothing}\right) = 0$

because $\mu$ is a measure.

$\Box$

Having verified a suitable set of axioms, it follows that $\mu \restriction_{\Sigma'}$ is a measure.

$\blacksquare$